AQA Further Paper 1 2023 June — Question 14 10 marks

Exam BoardAQA
ModuleFurther Paper 1 (Further Paper 1)
Year2023
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolar coordinates
TypeShow polar curve has Cartesian form
DifficultyChallenging +1.2 This is a structured Further Maths question on polar coordinates and conic sections. Part (a) requires finding the range by identifying max/min of the denominator (routine). Part (b) involves standard polar-to-Cartesian conversion using r²=x²+y² and x=r cos θ. Part (c) requires recognizing both curves as ellipses and identifying the transformation (translation and scaling). While this covers multiple techniques and requires careful algebra, each part follows well-established procedures without requiring novel insight—typical of a structured Further Maths question but more demanding than standard A-level due to the polar coordinate manipulation.
Spec4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta)

The curve C has polar equation $$r = \frac{A}{5 + 3 \cos \theta} \quad (-\pi < \theta \leq \pi)$$
  1. Show that \(r\) takes values in the range \(\frac{1}{k} \leq r \leq k\), where \(k\) is an integer. [2 marks]
  2. Find the Cartesian equation of C in the form \(y^2 = f(x)\) [4 marks]
  3. The ellipse E has equation $$y^2 + \frac{16x^2}{25} = 1$$ Find the transformation that maps the graph of E onto C [4 marks]

Question 14:

AnswerMarks
14(a)Uses the range of cosθ to
obtain max and min values of r
AnswerMarks Guidance
or obtains k =22.4 M1
and maximum when cosθ=−1
1
So ≤r ≤2
2
Completes a reasoned
argument to obtain the correct
AnswerMarks Guidance
inequality2.1 R1
Subtotal2
QMarking instructions AO

AnswerMarks Guidance
14(b)Uses x=rcosθ 3.1a
5r =4−3rcosθ
5 x2+ y2 =4−3x
25 ( x2+ y2) =(4−3x)2
25y2 =16−24x−16x2
y2 = 1 ( 16−24x−16x2)
25
r2 = x2 + y2
Uses
or
AnswerMarks Guidance
r = x2 + y23.1a M1
ky2
Rearranges to make the
subject, having correctly
removed any square root in their
equation. Equation must be in
AnswerMarks Guidance
terms of xand y only.1.1a M1
Obtains a correct result in
required form
AnswerMarks Guidance
ISW1.1b A1
Subtotal4
QMarking instructions AO

AnswerMarks Guidance
14(c)Completes the square to bring
together all the x terms3.1a M1
y2 =− x2+ x−1
25 2 
 2 
16 3 25
=− x+  − 
25  4 16
2
16 3
=− x+  +1
25 4
2
16 3
y2+ x+  =1
25 4
16x2
Starting from y2 + =1:
25
The required transformation is
a translation by  − 3
 
4
 
 0 
Rearranges equation to obtain a
form that enables them to
AnswerMarks Guidance
deduce a transformation1.1a M1
Deduces that a horizontal
translation is required.
p
PI a vector  
AnswerMarks Guidance
02.2a M1
Deduces a translation by  − 3
 
4
 
 0 
AnswerMarks Guidance
as the only transformation2.2a A1
Subtotal4
Question total10
QMarking instructions AO
Question 14:
--- 14(a) ---
14(a) | Uses the range of cosθ to
obtain max and min values of r
or obtains k =2 | 2.4 | M1 | r is minimum when cosθ=1 ,
and maximum when cosθ=−1
1
So ≤r ≤2
2
Completes a reasoned
argument to obtain the correct
inequality | 2.1 | R1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 14(b) ---
14(b) | Uses x=rcosθ | 3.1a | M1 | 5r+3rcosθ=4
5r =4−3rcosθ
5 x2+ y2 =4−3x
25 ( x2+ y2) =(4−3x)2
25y2 =16−24x−16x2
y2 = 1 ( 16−24x−16x2)
25
r2 = x2 + y2
Uses
or
r = x2 + y2 | 3.1a | M1
ky2
Rearranges to make the
subject, having correctly
removed any square root in their
equation. Equation must be in
terms of xand y only. | 1.1a | M1
Obtains a correct result in
required form
ISW | 1.1b | A1
Subtotal | 4
Q | Marking instructions | AO | Marks | Typical solution
--- 14(c) ---
14(c) | Completes the square to bring
together all the x terms | 3.1a | M1 | 16 3 
y2 =− x2+ x−1
25 2 
 2 
16 3 25
=− x+  − 
25  4 16
2
16 3
=− x+  +1
25 4
2
16 3
y2+ x+  =1
25 4
16x2
Starting from y2 + =1:
25
The required transformation is
a translation by  − 3
 
4
 
 0 
Rearranges equation to obtain a
form that enables them to
deduce a transformation | 1.1a | M1
Deduces that a horizontal
translation is required.
p
PI a vector  
0 | 2.2a | M1
Deduces a translation by  − 3
 
4
 
 0 
as the only transformation | 2.2a | A1
Subtotal | 4
Question total | 10
Q | Marking instructions | AO | Marks | Typical solution
The curve C has polar equation
$$r = \frac{A}{5 + 3 \cos \theta} \quad (-\pi < \theta \leq \pi)$$

\begin{enumerate}[label=(\alph*)]
\item Show that $r$ takes values in the range $\frac{1}{k} \leq r \leq k$, where $k$ is an integer.
[2 marks]

\item Find the Cartesian equation of C in the form $y^2 = f(x)$
[4 marks]

\item The ellipse E has equation
$$y^2 + \frac{16x^2}{25} = 1$$

Find the transformation that maps the graph of E onto C
[4 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 1 2023 Q14 [10]}}